Problem: Simplify the following expression and state the condition under which the simplification is valid. $t = \dfrac{4y^2 + 32y - 80}{-3y^2 + 3y + 6}$
Explanation: First factor out the greatest common factors in the numerator and in the denominator. $ t = \dfrac {4(y^2 + 8y - 20)} {-3(y^2 - y - 2)} $ $ t = -\dfrac{4}{3} \cdot \dfrac{y^2 + 8y - 20}{y^2 - y - 2} $ Next factor the numerator and denominator. $ t = - \dfrac{4}{3} \cdot \dfrac{(y - 2)(y + 10)}{(y - 2)(y + 1)}$ Assuming $y \neq 2$ , we can cancel the $y - 2$ $ t = - \dfrac{4}{3} \cdot \dfrac{y + 10}{y + 1}$ Therefore: $ t = \dfrac{ -4(y + 10)}{ 3(y + 1)}$, $y \neq 2$